Assuming Complete Dissociation, Calculate the Ph of the Following Solutions: 0.003 M Hcl - Chemistry
Assuming complete dissociation, calculate the pH of the following solutions:
0.003 M HCl
`H_2O + HCl ↔ H_3O^+ + Cl^(-)`
Since HCl is completely ionized,
`[H_3O^+] = [HCl]`
`=>[H_3O^+] = 0.003`
`pH = - log[H_3O^+] = -log(.003)`
Hence, the pH of the solution is 2.52.
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